Consider the following data for an unknown substance \(\mathrm{X} :\) $$\begin{array}{l}{\Delta H_{\mathrm{vap}}=20.00 \mathrm{kJ} / \mathrm{mol}} \\\ {\Delta H_{\mathrm{fus}}=5.00 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ $$\begin{array}{l}{\text { Specific heat capacity of solid }=3.00 \mathrm{Jg} \cdot^{\circ} \mathrm{C}} \\ {\text { Specific heat capacity of liquid }=2.50 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}} \\ {\text { Boiling point }=75.0^{\circ} \mathrm{C}} \\ {\text { Melting point }=-15.0^{\circ} \mathrm{C}} \\ {\text { Molar mass }=100.0 \mathrm{g} / \mathrm{mol}}\end{array}$$ In the heating of substance \(\mathrm{X}\) , energy (heat) is added at a constant rate of 450.0 \(\mathrm{J} / \mathrm{min}\) . At this rate, how long will it take to heat 10.0 \(\mathrm{g}\) of \(\mathrm{X}\) from $-35.0^{\circ} \mathrm{C}\( to \)25.0^{\circ} \mathrm{C} ?$

Step by step solution

01

Calculate heat required for heating the solid X from -35°C to -15°C

To calculate the heat required for this step, we can use the specific heat capacity of the solid along with the change in temperature: $$q_{1} = m C_{s} \Delta T$$ Where: \(q_{1}\) = the heat required in this step \(m\) = mass of substance X (10 g) \(C_{s}\) = specific heat capacity of solid X (3.00 J/g·°C) \(\Delta T\) = change in temperature (-15°C - (-35°C)) Plugging in the values, we get: $$q_{1} = (10 \, \text{g}) (3.00 \, \frac{\text{J}}{\text{g} \cdot ^{\circ} \text{C}}) (20 ^{\circ} \text{C}) = 600 \, \text{J}$$

02

Calculate heat required for melting the solid X at -15°C

To calculate the heat required for melting, we can use the enthalpy of fusion given in the exercise: $$q_{2} = m \Delta H_{\mathrm{fus}}$$ Where: \(q_{2}\) = the heat required in this step \(\Delta H_{\mathrm{fus}}\) = enthalpy of fusion (5.00 kJ/mol) First, we need to convert the mass of X to moles: $$n_\mathrm{X} = \frac{m}{M}$$ Where: \(n_\mathrm{X}\) = moles of substance X \(M\) = molar mass of X (100.0 g/mol) $$n_\mathrm{X} = \frac{10.0 \, \text{g}}{100.0 \, \frac{\text{g}}{\text{mol}}} = 0.1 \, \text{mol}$$ Now, we can find q2: $$q_{2} = (0.1 \, \text{mol}) (5000 \, \frac{\text{J}}{\text{mol}}) = 500 \, \text{J}$$

03

Calculate heat required for heating the liquid X from -15°C to 25°C

To calculate the heat required for this step, we can use the specific heat capacity of the liquid along with the change in temperature: $$q_{3} = m C_{l} \Delta T$$ Where: \(q_{3}\) = the heat required in this step \(C_{l}\) = specific heat capacity of liquid X (2.50 J/g·°C) \(\Delta T\) = change in temperature (25°C - (-15°C)) Plugging in the values, we get: $$q_{3} = (10\, \text{g}) (2.50 \, \frac{\text{J}}{\text{g} \cdot ^{\circ} \text{C}}) (40 ^{\circ}\text{C}) = 1000 \, \text{J}$$

04

Calculate the total heat required

Now, we can add up all the heat required in each step: $$q_\text{total} = q_{1} + q_{2} + q_{3} = 600 \, \text{J} + 500 \, \text{J} + 1000 \, \text{J} = 2100 \, \text{J}$$

05

Calculate the time required

Finally, we can use the heat transfer rate given in the exercise to calculate the time required to heat the 10 g of substance X: $$t = \frac{q_\text{total}}{\text{rate}}$$ Where: \(t\) = time required \(\text{rate}\) = heat transfer rate (450 J/min) $$t = \frac{2100 \, \text{J}}{450 \, \frac{\text{J}}{\text{min}}} = 4.67 \, \text{min}$$ Therefore, it will take approximately 4.67 minutes to heat 10.0 g of substance X from -35.0°C to 25.0°C at the given heat transfer rate.

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